$$r = \frac{3}{4} \qquad an = \frac{27}{4} \qquad Sn = 43.75 \qquad $$ Find $n$ and $a\tiny{1} $ of G.P
I tried with the $Sn$ formula but it takes so much time and I am not going to the answer,in the end, I have to put numbers in n from 2, to get the right answer,and it is too much work for this kind of problem.
Suppose $a_1, a_2, \ldots, a_n$ are in geometric progression with common ratio $r$. Then from the end, the sequence $$a_n, a_{n-1}, \ldots, a_2, a_1$$
is also a geometric progression with common ratio $1/r$.
So treat $27/4$ as first term, $4/3$ as common ratio and given sum is $175/4$. Now apply the usual sum formula $$\frac{175}{4} = \frac{27}{4}\times \frac{(4/3)^n-1}{(4/3)-1}$$
to compute $n$. Then you can also find $a_1$ as last term of our new sequence. $$a_1 = \frac{27}{4} \times \Big( \frac{4}{3} \Big)^{n-1}$$